Linear Algebra: Spaces and Maps

Y. Eddie Lu, Summer 2026

ECON 8001 course index · Lecture 2 of 8

Orientation

What problem does this solve?

Many economic objects can be combined and rescaled: portfolio positions, payoff vectors, factor exposures, perturbations of a parameter, and local changes in an equilibrium condition. Linear algebra asks which directions are available and what a transformation preserves or destroys.

The central picture

A vector space is a collection of feasible directions through an origin. A linear map takes directions in one space to directions in another while respecting combinations. Its kernel records directions that disappear; its image records output directions that can actually be reached.

Where this sits

This page uses the logic and function language from Foundations. It supplies the coordinate-free objects that matrices represent on the next linear-algebra page. It also prepares the Jacobian, Hessian, and tangent-space arguments later in the course.

Prerequisite retrieval

Retrieval check. Why is \(\{(x,y)^\top\in\mathbb R^2:x+y=1\}\) not a vector subspace of \(\mathbb R^2\)?

Show answer

It does not contain the zero vector: \(0+0\ne1\). It is an affine line, not a subspace. Closure under addition also fails, since \((1,0)^\top\) added to itself is not in the set.

Vector spaces and subspaces

Definition: vector space

Let \(\mathbb F\in\{\mathbb R,\mathbb C\}\). A vector space over \(\mathbb F\) is a set \(V\) with operations

\[ \begin{aligned} +&:V\times V\to V,\\ \cdot&:\mathbb F\times V\to V. \end{aligned} \]

such that, for every \(u,v,w\in V\) and \(r,s\in\mathbb F\):

\[ \begin{aligned} u+v&=v+u,\\ (u+v)+w&=u+(v+w),\\ \exists 0\in V:\quad v+0&=v,\\ \exists(-v)\in V:\quad v+(-v)&=0,\\ r(u+v)&=ru+rv,\\ (r+s)v&=rv+sv,\\ r(sv)&=(rs)v,\\ 1v&=v. \end{aligned} \]

The operations must also be closed: \(u+v\in V\) and \(rv\in V\).

The axioms say that we may add feasible directions, undo a direction by taking its negative, and scale a direction without leaving the space.

Example: portfolios with unrestricted positions

The set \(\mathbb R^n\) of column vectors is a vector space over \(\mathbb R\). If \(x,y\in\mathbb R^n\) are portfolios and \(a,b\in\mathbb R\), then \(ax+by\in\mathbb R^n\). Short positions are what make multiplication by a negative scalar feasible.

Nonexample: a no-short-sale cone

The set \(\mathbb R^n_+:=\{x\in\mathbb R^n:x\geq0\}\) is not a vector space when \(n\geq1\). It is closed under addition and multiplication by nonnegative scalars, but \(x\in\mathbb R^n_+\) with \(x\ne0\) implies \(-x\notin\mathbb R^n_+\).

Definition: vector subspace

If \(V\) is a vector space, a subset \(U\subseteq V\) is a vector subspace when \(U\), with the operations inherited from \(V\), is itself a vector space.

Subspace test

Assumptions. \(V\) is a vector space over \(\mathbb F\), and \(U\subseteq V\) is nonempty.

Conclusion. \(U\) is a vector subspace of \(V\) if and only if

\[ \forall u,v\in U\;\forall a,b\in\mathbb F:\quad au+bv\in U. \]

Show proof

Proof strategy: reduce ten axioms to closure

The ambient space \(V\) already supplies associativity, commutativity, distributivity, and scalar identities. For \(U\), it remains to ensure that the inherited operations never leave \(U\). Nonemptiness plus closure under all linear combinations gives the zero vector and additive inverses automatically.

Proof. If \(U\) is a subspace, closure under \(au+bv\) is immediate. Conversely, choose \(u_0\in U\), which exists by nonemptiness. Then \(0=0u_0+0u_0\in U\). For each \(u\in U\), \(-u=(-1)u+0\in U\). Taking \(a=b=1\) gives closure under addition, and taking \(b=0\) gives closure under scalar multiplication. The remaining axioms are inherited from \(V\). \(\square\)

Example: a homogeneous restriction

Let \(U:=\{(x,y,z)^\top\in\mathbb R^3:x+y+z=0\}\). If \(u,v\in U\) and \(a,b\in\mathbb R\), then the coordinates of \(au+bv\) sum to \(a\cdot0+b\cdot0=0\). Thus \(U\) is a subspace. Homogeneous linear restrictions typically define subspaces.

Boundary case: an affine restriction

Replacing the right-hand side by \(1\) creates \(\{(x,y,z)^\top:x+y+z=1\}\). The zero vector fails the restriction, so the set is not a subspace. It is a translate of a subspace, called an affine set.

Span, independence, basis, and dimension

Definition: span

For vectors \(v_1,\ldots,v_k\in V\),

\[ \operatorname{span}(v_1,\ldots,v_k) \]

is the set of all vectors of the form

\[ \sum_{j=1}^k a_jv_j, \]

where \(a_1,\ldots,a_k\in\mathbb F\).

It is the smallest vector subspace of \(V\) containing every \(v_j\).

Span means reachable directions

One nonzero vector in \(\mathbb R^2\) spans a line through the origin. Two vectors that do not point along the same line span the plane. Span answers whether a proposed collection is sufficient to construct the vectors of interest.

Definition: linear independence

The list \(v_1,\ldots,v_k\in V\) is linearly independent if

\[ \sum_{j=1}^k a_jv_j=0 \quad\Longrightarrow\quad a_1=\cdots=a_k=0. \]

It is linearly dependent if there is a solution to this equation with at least one \(a_j\ne0\).

Independence means that no listed direction can be generated from the others. Dependence does not prevent a list from spanning a space. It tells you that the spanning list includes a redundant vector.

Example: dependent but spanning

The list \((1,0)^\top,(0,1)^\top,(1,1)^\top\) spans \(\mathbb R^2\), because the first two vectors do. It is dependent because

\[ (1,0)^\top+(0,1)^\top-(1,1)^\top=0. \]

Definition: basis and dimension

A list \(v_1,\ldots,v_k\) is a basis of \(V\) if it spans \(V\) and is linearly independent. A vector space is finite-dimensional if it has a finite basis. Its dimension, denoted \(\dim V\), is the number of vectors in any basis of \(V\).

Coordinates relative to a basis

Assumptions. \(v_1,\ldots,v_k\) is a basis of \(V\).

Conclusion. Every \(v\in V\) has a unique coordinate representation

\[ v=a_1v_1+\cdots+a_kv_k \]

for scalars \(a_1,\ldots,a_k\in\mathbb F\).

Show proof

Proof strategy: use the two parts of a basis separately

Spanning gives existence of a representation. Independence gives uniqueness, because the difference of two representations is a linear combination equal to zero.

Proof. Since the list spans \(V\), at least one representation exists. If both \(\sum_j a_jv_j\) and \(\sum_j b_jv_j\) equal \(v\), subtracting gives \(\sum_j(a_j-b_j)v_j=0\). Linear independence implies \(a_j-b_j=0\) for all \(j\), so \(a_j=b_j\) for all \(j\). \(\square\)

Worked example: a basis for a constrained plane

For \(U=\{(x,y,z)^\top\in\mathbb R^3:x+y+z=0\}\),

\[ (x,y,z)^\top=x(1,0,-1)^\top+y(0,1,-1)^\top. \]

The two displayed vectors span \(U\). If \(a(1,0,-1)^\top+b(0,1,-1)^\top=0\), the first two coordinates give \(a=b=0\), so they are independent. Therefore they form a basis and \(\dim U=2\).

Retrieval check. What does each half of the word “basis” contribute: span and independence?

Show answer

Span gives existence of coordinates: every vector can be represented. Independence gives uniqueness: no vector has two different coordinate lists relative to the basis.

Linear maps

Definition: linear map

Let \(V\) and \(W\) be vector spaces over the same field \(\mathbb F\). A function \(L:V\to W\) is linear if, for every \(u,v\in V\) and \(a,b\in\mathbb F\),

\[ L(au+bv)=aL(u)+bL(v). \]

Equivalently, \(L(u+v)=L(u)+L(v)\) and \(L(av)=aL(v)\) hold for all permitted inputs and scalars.

Every linear map satisfies \(L(0)=0\). Indeed, \(L(0)=L(0+0)=L(0)+L(0)\), and additive cancellation in \(W\) gives the result.

Example: a factor-exposure map

Let \(B\in\mathbb R^{m\times n}\) and define \(L:\mathbb R^n\to\mathbb R^m\) by \(L(x)=Bx\). Here \(x\) is an \(n\times1\) position vector and \(Bx\) is an \(m\times1\) vector of factor exposures. For \(a,b\in\mathbb R\) and \(x,y\in\mathbb R^n\), matrix distributivity gives \(B(ax+by)=aBx+bBy\), so \(L\) is linear.

Nonexample: an intercept destroys linearity

The affine map \(f:\mathbb R^n\to\mathbb R^m\) given by \(f(x)=Bx+c\) is not linear when \(c\ne0\), because \(f(0)=c\ne0\). Its derivative is the linear map \(x\mapsto Bx\), but the function itself is not linear.

The space of linear maps

Assumptions. \(V\) and \(W\) are vector spaces over \(\mathbb F\). Let \(\mathcal L(V,W)\) be the set of linear maps from \(V\) to \(W\). For \(L_1,L_2\in\mathcal L(V,W)\) and \(a\in\mathbb F\), define

\[ \begin{aligned} (L_1+L_2)(v)&:=L_1(v)+L_2(v),\\ (aL_1)(v)&:=aL_1(v). \end{aligned} \]

Conclusion. With these pointwise operations, \(\mathcal L(V,W)\) is a vector space.

The important operational point is closure: a linear combination of linear maps is again linear. This permits one to treat a parameterized family of maps as an ordinary vector space.

Kernel, image, and rank-nullity

Definition: kernel and image

For a linear map \(L:V\to W\),

\[ \begin{aligned} \ker L&:=\{v\in V:L(v)=0\},\\ \operatorname{im}L&:=\{L(v):v\in V\}\subseteq W. \end{aligned} \]

The kernel is a subset of the domain \(V\). The image is a subset of the codomain \(W\).

Information accounting

If two inputs differ by a vector in \(\ker L\), then they have the same output. Thus a nonzero kernel records lost distinctions. The image records the output directions that remain available after the transformation.

Kernel and image are subspaces

Assumptions. \(L:V\to W\) is a linear map.

Conclusion. \(\ker L\) is a vector subspace of \(V\), and \(\operatorname{im}L\) is a vector subspace of \(W\).

Show proof

Proof strategy: apply the subspace test and linearity

For the kernel, show that a linear combination of inputs mapped to zero is again mapped to zero. For the image, write two output vectors as images of inputs and combine those inputs before applying \(L\).

Proof. If \(u,v\in\ker L\) and \(a,b\in\mathbb F\), then \(L(au+bv)=aL(u)+bL(v)=0\), so \(au+bv\in\ker L\). If \(y_1,y_2\in\operatorname{im}L\), choose \(u,v\in V\) with \(y_1=L(u)\) and \(y_2=L(v)\). Then \(ay_1+by_2=L(au+bv)\in\operatorname{im}L\). The subspace test applies in both cases. \(\square\)

Rank-nullity theorem

Assumptions. \(V\) is finite-dimensional, \(W\) is a vector space over the same field, and \(L:V\to W\) is linear.

Conclusion. \(\operatorname{im}L\) is finite-dimensional and

\[ \dim V=\dim(\ker L)+\dim(\operatorname{im}L). \]

The quantity \(\dim(\operatorname{im}L)\) is the rank of \(L\); \(\dim(\ker L)\) is its nullity.

Show proof

Proof strategy: extend a kernel basis

Start with a basis of \(\ker L\). Extend it to a basis of \(V\). Under \(L\), the added basis vectors become a basis of the image: they span because the kernel coordinates disappear, and they are independent because a zero combination would place their preimage in the kernel. Counting the two blocks yields the identity.

Worked example: a map from \(\mathbb R^3\) to \(\mathbb R^2\)

Let \(L:\mathbb R^3\to\mathbb R^2\) be

\[ L(x,y,z)^\top=(x+y,z)^\top. \]

The equation \(L(x,y,z)^\top=0\) gives \(x=-y\) and \(z=0\), so

\[ \ker L=\operatorname{span}\{(1,-1,0)^\top\}. \]

Given any \((a,b)^\top\in\mathbb R^2\), choose \((x,y,z)^\top=(a,0,b)^\top\). Hence \(\operatorname{im}L=\mathbb R^2\). The dimensions satisfy \(3=1+2\).

Injective, surjective, and bijective maps

Definition: injective and surjective

A function \(f:X\to Y\) is injective if

\[ \forall x_1,x_2\in X:\quad f(x_1)=f(x_2)\Rightarrow x_1=x_2. \]

It is surjective if \(f(X)=Y\). It is bijective if it is both injective and surjective.

Injectivity and the kernel

Assumptions. \(L:V\to W\) is linear.

Conclusion. \(L\) is injective if and only if \(\ker L=\{0\}\).

Show proof

Proof. If \(L\) is injective and \(v\in\ker L\), then \(L(v)=0=L(0)\), so \(v=0\). Conversely, suppose \(\ker L=\{0\}\) and \(L(u)=L(v)\). Linearity gives \(L(u-v)=0\), so \(u-v=0\) and \(u=v\). \(\square\)

Assumption audit: codomain versus image

For \(L:V\to W\), surjectivity is the equality \(\operatorname{im}L=W\). A map may have a perfectly well-defined codomain much larger than its image. Do not label it surjective merely because the formula produces outputs in \(W\).

Dimension constraints

Assumptions. \(V\) and \(W\) are finite-dimensional vector spaces and \(L:V\to W\) is linear.

Conclusion. If \(\dim V>\dim W\), then \(L\) cannot be injective. If \(\dim V<\dim W\), then \(L\) cannot be surjective.

For the first claim, rank-nullity implies \(\dim\ker L\geq\dim V-\dim W>0\). For the second, \(\dim\operatorname{im}L\leq\dim V<\dim W\), so the image cannot equal \(W\).

Invalid inference: dimensions alone do not guarantee surjectivity

When \(\dim V=\dim W\), a linear map may still fail to be injective and surjective. The zero map from \(\mathbb R^2\) to \(\mathbb R^2\) has equal domain and codomain dimensions but rank zero. Equality of dimensions makes injectivity and surjectivity equivalent for a particular linear map; it does not make either automatic.

Economic and econometric use

In a linear regression, the column space of a regressor matrix contains fitted-value directions. Its null space contains coefficient perturbations that do not alter fitted values. Full column rank removes this ambiguity and is an identification condition for the coefficient vector. In constrained optimization, a tangent space is a subspace of feasible local directions; the second-order condition examines a quadratic form only along those directions.

Calculation blueprint

For a linear map \(L:V\to W\):

1. State the domain and codomain, including dimensions in coordinate spaces.
2. To find \(\ker L\), solve \(L(v)=0\) and express all solutions as a span.
3. To find \(\operatorname{im}L\), characterize all reachable outputs or span the images of a basis.
4. Check rank-nullity when \(V\) is finite-dimensional.
5. State injectivity and surjectivity as conclusions about the kernel and image, not as guesses from the formula.

Exit tickets

1. Is \(U=\{(x,y)^\top\in\mathbb R^2:2x-y=0\}\) a subspace? Give the shortest valid proof.
2. Let \(L:\mathbb R^3\to\mathbb R^2\) satisfy \(L(x,y,z)^\top=(x-z,y-z)^\top\). Find a basis for \(\ker L\), determine whether \(L\) is surjective, and verify rank-nullity.
3. State the exact difference between an injective map and a surjective map without using an example.

Show answer

1. Yes. It is nonempty and, if \(2x-y=0\) and \(2x'-y'=0\), then \(2(ax+bx')-(ay+by')=0\) for all \(a,b\in\mathbb R\). The subspace test applies.

2. The equations \(x-z=0\) and \(y-z=0\) give \(x=y=z\), so \(\ker L=\operatorname{span}\{(1,1,1)^\top\}\). The map is surjective because \((a,b)^\top=L(a,b,0)^\top\) for every \((a,b)^\top\in\mathbb R^2\). Thus \(3=1+2\).

3. Injective means equal outputs imply equal inputs. Surjective means every element of the declared codomain has at least one preimage.

Mastery check

You should now be able to apply the subspace test, separate span from independence, prove coordinate uniqueness from a basis, define kernel and image with the correct ambient spaces, and use rank-nullity to audit a dimension claim. The next page will express these objects in coordinates and matrices.